let t1 = Node ((Node ((Leaf 2), (Leaf 3))), (Node ((Leaf 5),(Node ((Leaf 7), (Leaf 11)))))) . ___|___ | | . . _|__ _|__ | | | | 2 3 5 . _|__ | | 7 11Our first task will be to replace each leaf with its double:

let rec treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) = match t with Leaf x -> Leaf (newleaf x) | Node (l, r) -> Node ((treemap newleaf l), (treemap newleaf r));;`treemap` takes a function that transforms old leaves into new leaves, and maps that function over all the leaves in the tree, leaving the structure of the tree unchanged. For instance:

let double i = i + i;; treemap double t1;; - : int tree = Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22))) . ___|____ | | . . _|__ __|__ | | | | 4 6 10 . __|___ | | 14 22We could have built the doubling operation right into the `treemap` code. However, because what to do to each leaf is a parameter, we can decide to do something else to the leaves without needing to rewrite `treemap`. For instance, we can easily square each leaf instead by supplying the appropriate `int -> int` operation in place of `double`:

let square x = x * x;; treemap square t1;; - : int tree =ppp Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))Note that what `treemap` does is take some global, contextual information---what to do to each leaf---and supplies that information to each subpart of the computation. In other words, `treemap` has the behavior of a reader monad. Let's make that explicit. In general, we're on a journey of making our treemap function more and more flexible. So the next step---combining the tree transducer with a reader monad---is to have the treemap function return a (monadized) tree that is ready to accept any `int->int` function and produce the updated tree. \tree (. (. (f2) (f3))(. (f5) (.(f7)(f11))))

\f . ____|____ | | . . __|__ __|__ | | | | f2 f3 f5 . __|___ | | f7 f11That is, we want to transform the ordinary tree `t1` (of type `int tree`) into a reader object of type `(int->int)-> int tree`: something that, when you apply it to an `int->int` function returns an `int tree` in which each leaf `x` has been replaced with `(f x)`. With previous readers, we always knew which kind of environment to expect: either an assignment function (the original calculator simulation), a world (the intensionality monad), an integer (the Jacobson-inspired link monad), etc. In this situation, it will be enough for now to expect that our reader will expect a function of type `int->int`.

type 'a reader = (int->int) -> 'a;; (* mnemonic: e for environment *) let reader_unit (x:'a): 'a reader = fun _ -> x;; let reader_bind (u: 'a reader) (f:'a -> 'c reader):'c reader = fun e -> f (u e) e;;It's easy to figure out how to turn an `int` into an `int reader`:

let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;; int2int_reader 2 (fun i -> i + i);; - : int = 4But what do we do when the integers are scattered over the leaves of a tree? A binary tree is not the kind of thing that we can apply a function of type `int->int` to.

let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader = match t with Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x')) | Node (l, r) -> reader_bind (treemonadizer f l) (fun x -> reader_bind (treemonadizer f r) (fun y -> reader_unit (Node (x, y))));;This function says: give me a function `f` that knows how to turn something of type `'a` into an `'b reader`, and I'll show you how to turn an `'a tree` into an `'a tree reader`. In more fanciful terms, the `treemonadizer` function builds plumbing that connects all of the leaves of a tree into one connected monadic network; it threads the monad through the leaves.

# treemonadizer int2int_reader t1 (fun i -> i + i);; - : int tree = Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))Here, our environment is the doubling function (`fun i -> i + i`). If we apply the very same `int tree reader` (namely, `treemonadizer int2int_reader t1`) to a different `int->int` function---say, the squaring function, `fun i -> i * i`---we get an entirely different result:

# treemonadizer int2int_reader t1 (fun i -> i * i);; - : int tree = Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))Now that we have a tree transducer that accepts a monad as a parameter, we can see what it would take to swap in a different monad. For instance, we can use a state monad to count the number of nodes in the tree.

type 'a state = int -> 'a * int;; let state_unit x i = (x, i+.5);; let state_bind u f i = let (a, i') = u i in f a (i'+.5);;Gratifyingly, we can use the `treemonadizer` function without any modification whatsoever, except for replacing the (parametric) type `reader` with `state`:

let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state = match t with Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x')) | Node (l, r) -> state_bind (treemonadizer f l) (fun x -> state_bind (treemonadizer f r) (fun y -> state_unit (Node (x, y))));;Then we can count the number of nodes in the tree:

# treemonadizer state_unit t1 0;; - : int tree * int = (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13) . ___|___ | | . . _|__ _|__ | | | | 2 3 5 . _|__ | | 7 11Notice that we've counted each internal node twice---it's a good exercise to adjust the code to count each node once. One more revealing example before getting down to business: replacing `state` everywhere in `treemonadizer` with `list` gives us

# treemonadizer (fun x -> [ [x; square x] ]) t1;; - : int list tree list = [Node (Node (Leaf [2; 4], Leaf [3; 9]), Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]Unlike the previous cases, instead of turning a tree into a function from some input to a result, this transformer replaces each `int` with a list of `int`'s. Now for the main point. What if we wanted to convert a tree to a list of leaves?

type ('a, 'r) continuation = ('a -> 'r) -> 'r;; let continuation_unit x c = c x;; let continuation_bind u f c = u (fun a -> f a c);; let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation = match t with Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x')) | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x -> continuation_bind (treemonadizer f r) (fun y -> continuation_unit (Node (x, y))));;We use the continuation monad described above, and insert the `continuation` type in the appropriate place in the `treemonadizer` code. We then compute:

# treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);; - : int list = [2; 3; 5; 7; 11]We have found a way of collapsing a tree into a list of its leaves. The continuation monad is amazingly flexible; we can use it to simulate some of the computations performed above. To see how, first note that an interestingly uninteresting thing happens if we use the continuation unit as our first argument to `treemonadizer`, and then apply the result to the identity function:

# treemonadizer continuation_unit t1 (fun x -> x);; - : int tree = Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))That is, nothing happens. But we can begin to substitute more interesting functions for the first argument of `treemonadizer`:

(* Simulating the tree reader: distributing a operation over the leaves *) # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);; - : int tree = Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121))) (* Simulating the int list tree list *) # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);; - : int list tree = Node (Node (Leaf [2; 4], Leaf [3; 9]), Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121]))) (* Counting leaves *) # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);; - : int = 5We could simulate the tree state example too, but it would require generalizing the type of the continuation monad to type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;; The binary tree monad --------------------- Of course, by now you may have realized that we have discovered a new monad, the binary tree monad:

type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);; let tree_unit (x:'a) = Leaf x;; let rec tree_bind (u:'a tree) (f:'a -> 'b tree):'b tree = match u with Leaf x -> f x | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;For once, let's check the Monad laws. The left identity law is easy: Left identity: bind (unit a) f = bind (Leaf a) f = fa To check the other two laws, we need to make the following observation: it is easy to prove based on `tree_bind` by a simple induction on the structure of the first argument that the tree resulting from `bind u f` is a tree with the same strucure as `u`, except that each leaf `a` has been replaced with `fa`: \tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))

. . __|__ __|__ | | | | a1 . fa1 . _|__ __|__ | | | | . a5 . fa5 bind _|__ f = __|__ | | | | . a4 . fa4 __|__ __|___ | | | | a2 a3 fa2 fa3Given this equivalence, the right identity law Right identity: bind u unit = u falls out once we realize that bind (Leaf a) unit = unit a = Leaf a As for the associative law, Associativity: bind (bind u f) g = bind u (\a. bind (fa) g) we'll give an example that will show how an inductive proof would proceed. Let `f a = Node (Leaf a, Leaf a)`. Then \tree (. (. (. (. (a1)(a2))))) \tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) ))

. ____|____ . . | | bind __|__ f = __|_ = . . | | | | __|__ __|__ a1 a2 fa1 fa2 | | | | a1 a1 a1 a1Now when we bind this tree to `g`, we get

. ____|____ | | . . __|__ __|__ | | | | ga1 ga1 ga1 ga1At this point, it should be easy to convince yourself that using the recipe on the right hand side of the associative law will built the exact same final tree. So binary trees are a monad. Haskell combines this monad with the Option monad to provide a monad called a [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree) that is intended to represent non-deterministic computations as a tree. Refunctionalizing zippers: from lists to continuations ------------------------------------------------------ Let's work with lists of chars for a change. To maximize readability, we'll indulge in an abbreviatory convention that "abc" abbreviates the list `['a'; 'b'; 'c']`. Task 1: replace each occurrence of 'S' with a copy of the string up to that point. Expected behavior:

t1 "abSe" ~~> "ababe"In linguistic terms, this is a kind of anaphora resolution, where `'S'` is functioning like an anaphoric element, and the preceding string portion is the antecedent. This deceptively simple task gives rise to some mind-bending complexity. Note that it matters which 'S' you target first (the position of the * indicates the targeted 'S'):

t1 "aSbS" * ~~> t1 "aabS" * ~~> "aabaab"versus

t1 "aSbS" * ~~> t1 "aSbaSb" * ~~> t1 "aabaSb" * ~~> "aabaaabab"versus

t1 "aSbS" * ~~> t1 "aSbaSb" * ~~> t1 "aSbaaSbab" * ~~> t1 "aSbaaaSbaabab" * ~~> ...Aparently, this task, as simple as it is, is a form of computation, and the order in which the `'S'`s get evaluated can lead to divergent behavior. For now, as usual, we'll agree to always evaluate the leftmost `'S'`. This is a task well-suited to using a zipper.

type 'a list_zipper = ('a list) * ('a list);; let rec t1 (z:char list_zipper) = match z with (sofar, []) -> List.rev(sofar) (* Done! *) | (sofar, 'S'::rest) -> t1 ((List.append sofar sofar), rest) | (sofar, fst::rest) -> t1 (fst::sofar, rest);; (* Move zipper *) # t1 ([], ['a'; 'b'; 'S'; 'e']);; - : char list = ['a'; 'b'; 'a'; 'b'; 'e'] # t1 ([], ['a'; 'S'; 'b'; 'S']);; - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']Note that this implementation enforces the evaluate-leftmost rule. Task 1 completed. One way to see exactly what is going on is to watch the zipper in action by tracing the execution of `t1`. By using the `#trace` directive in the Ocaml interpreter, the system will print out the arguments to `t1` each time it is (recurcively) called:

# #trace t1;; t1 is now traced. # t1 ([], ['a'; 'b'; 'S'; 'e']);; t1 <-- ([], ['a'; 'b'; 'S'; 'e']) t1 <-- (['a'], ['b'; 'S'; 'e']) t1 <-- (['b'; 'a'], ['S'; 'e']) t1 <-- (['b'; 'a'; 'b'; 'a'], ['e']) t1 <-- (['e'; 'b'; 'a'; 'b'; 'a'], []) t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] - : char list = ['a'; 'b'; 'a'; 'b'; 'e']The nice thing about computations involving lists is that it's so easy to visualize them as a data structure. Eventually, we want to get to a place where we can talk about more abstract computations. In order to get there, we'll first do the exact same thing we just did with concrete zipper using procedures. Think of a list as a procedural recipe: `['a'; 'b'; 'c']` means (1) start with the empty list `[]`; (2) make a new list whose first element is 'c' and whose tail is the list construted in the previous step; (3) make a new list whose first element is 'b' and whose tail is the list constructed in the previous step; and (4) make a new list whose first element is 'a' and whose tail is the list constructed in the previous step. What is the type of each of these steps? Well, it will be a function from the result of the previous step (a list) to a new list: it will be a function of type `char list -> char list`. We'll call each step a **continuation** of the recipe. So in this context, a continuation is a function of type `char list -> char list`. This means that we can now represent the sofar part of our zipper--the part we've already unzipped--as a continuation, a function describing how to finish building the list:

let rec t1c (l: char list) (c: (char list) -> (char list)) = match l with [] -> c [] | 'S'::rest -> t1c rest (fun x -> c (c x)) | a::rest -> t1c rest (fun x -> List.append (c x) [a]);; # t1c ['a'; 'b'; 'S'] (fun x -> x);; - : char list = ['a'; 'b'; 'a'; 'b'] # t1c ['a'; 'S'; 'b'; 'S'] (fun x -> x);; - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']Note that we don't need to do any reversing.